# How do you solve the exponential equation #2^(x+1)=16^(x+2)#?

##### 1 Answer

Real solution:

#x = -7/3#

Complex solutions:

#x = -7/3+(2kpii)/(3 ln 2)" "# for any#k in ZZ#

#### Explanation:

If

#(a^b)^c = a^(bc)#

So:

#2^(x+1) = 16^(x+2) = (2^4)^(x+2) = 2^(4(x+2))#

The function

#2^a = 2^b" "<=>" "a = b#

So in our example, we must have:

#x+1 = 4(x+2) = 4x+8#

Subtract

#-7 = 3x#

Divide both sides by

#x = -7/3#

**Complex solutions**

Note that:

#e^(2kpii) = 1#

for any integer value of

Hence:

#2^((2kpii)/ln 2) = e^(ln 2 * ((2kpii)/ln 2)) = e^(2kpii) = 1#

Hence for Complex values of

#2^a = 2^b" "<=>" "a = b+(2kpii)/ln 2 " for some " k in ZZ#

So in our example we must have:

#4x + 8 = x + 1 + (2kpii)/ln 2#

Subtract

#3x = -7+(2kpii)/ln 2#

Divide both sides by

#x = -7/3+(2kpii)/(3 ln 2)#